Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to understand the axioms of axiomatic set theory. I'm studying this book and I didn't understand how can we define the elements of a set and the set $\{x\}$. If I define the singleton, I know how to define the finite sets using the union axiom.

Afterwards, the author says

The author just says: "given any $x$ we have the singleton $\{x\}$", which axiom he used to say that? afterwards the author says the definition of $\{x\}$, which in my opinion is wrong because using (I), $\{x\}$ should be define as $\{\{x\},\{x\}\}$.

I need help

Thanks in advance

share|improve this question
1  
See page 9 of your book : "Our theory will ignore all objects that are not sets." –  Mauro ALLEGRANZA Apr 10 at 15:49
    
@MauroALLEGRANZA Thank you for point this out! –  user42912 Apr 10 at 19:13

3 Answers 3

Uses the pair set axiom $\{x\}$ is by definition $\{x,x\}$.

"For any sets $u$ and $v$ ($x$ and $x$) the pair set $\{u,v\}$ ($\{x,x\}$) is the set...".

share|improve this answer
    
As I said in my question, in the pairing axiom $u$ and $v$ are sets, so using $u=v=\{x\}$ we should have $\{\{x\},\{x\}\}$, instead of $\{x,x\}$ –  user42912 Apr 10 at 13:43
1  
Using $u=v=x$.... –  Martín-Blas Pérez Pinilla Apr 10 at 13:49
    
But the author says $u$ and $v$ has to be sets. $x$ is an element not a set. –  user42912 Apr 10 at 13:49
1  
In the usual (without urelements) axiomatic set theory, all is a set. –  Martín-Blas Pérez Pinilla Apr 10 at 13:50
    
So if "all" is a set, what's the meaning of $x\in B$? –  user42912 Apr 10 at 15:58

In the Pairing Axiom, there is nothing requiring the variables $u$ and $v$ to take distinct values. In the case $u=v$, the resulting set $B$ is a singleton set.

share|improve this answer
    
As I said in my question, in the pairing axiom $u$ and $v$ are sets, so using $u=v=\{x\}$ we should have $\{\{x\},\{x\}\}$, instead of $\{x,x\}$ –  user42912 Apr 10 at 13:44
1  
@user42912: No, don't set $u=v=\{x\}$. Set $u=v=x$. Then $\{u,v\}=\{x,x\}=\{x\}$. –  Henning Makholm Apr 10 at 13:46
    
But the author says $u$ and $v$ has to be sets. $x$ is an element not a set. –  user42912 Apr 10 at 13:49
1  
@user42912: When you do standard axiomatic set theory, everything you speak of is a set. There's no such thing in the theory as an "element that is not a set". –  Henning Makholm Apr 10 at 13:51
    
So if everything I speak is a set, what's the meaning of $x\in B$? –  user42912 Apr 10 at 15:58

In axiomatic set theory we don't "define elements". Everything1 is an element, and everything is a set. There is no issue with sets being elements of other sets, if you consider the power set of $A$, it's a set and all its elements are sets themselves.

Axiomatic set theory says what are the properties of the $\in$ relation, and what sets we can prove to exist (given the existence of some sets).

For example, the axiom of pairing says that if $x$ and $y$ exist, then the set $\{x,y\}$ exists. Since we do not require that $x$ and $y$ are distinct, if $x=y$ we have that $\{x,x\}$ exists. But since repetition is discarded when considering sets, we have that $\{x,x\}=\{x\}$.

It might be strange, at first, to talk about "proving that $\{x,y\}$ exists", because when we first start with mathematics we are used to be given a world, and work in that world. The real numbers are given, we don't have to prove that $\sqrt2$ or $\pi$ exist, of course they do. There's no question.

But in logic, and in set theory, we transcend to a more abstract universe of mathematics, a universe which are we not familiar with, and can't fully imagine. And in that universe all we have is the axioms of set theory and $\in$, to light up our way as we trudge through the sets. So if you want to argue that there is a particular set, e.g. $\{x\}$, in the universe of set theory, then you have to prove that from your axioms.

Even if it seems "obvious", and even if it seems "preposterous". All that matters is that we can prove or disprove things. Luckily the axioms were chosen to be very compatible with our intuition when it comes to basic things like the existence of $\{x\}$, or the power set of $x$, and so on.


Footnotes.

  1. There are set theories in which we allow the existence of proper classes, there sets are defined as classes which are elements of other classes.

    There are set theories in which we allow the existence of atoms, or urelements, which are not sets, there sets are defined as objects which have elements (and the empty set as well).

share|improve this answer
    
Thank you very much Asaf for your answer, I don't have problems with axiomatic mathematics, I've already had a course in euclidian geometry and we studied it in an axiomatic way. My only problem is this statement: "everything is a set", I was used to a world with elements and sets. Of course there are sets of sets, but soon or later we have to have elements in a set at a certain point, that was my world until today. –  user42912 Apr 10 at 19:10
1  
Well. No more. You are free little bird, fly into the set filled skies!! Or something... :-) –  Asaf Karagila Apr 10 at 20:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.